Statistics-And-Probability Question 609
Question: Mean of the numbers $ 1,2,3,…,n $ with respective weights $ 1^{2}+1,2^{2}+2,3^{3}+3,…n^{2}+n $ is
Options:
A) $ \frac{3n(n+1)}{2(2n+1)} $
B) $ \frac{2n+1}{3} $
C) $ \frac{3n+1}{4} $
D) $ \frac{3n+1}{2} $
Show Answer
Answer:
Correct Answer: C
Solution:
[c] Here for each $ x_{i}=i, $ weight $ w_{i}=i^{2}+i $ Hence, the required mean $ =\frac{\sum{w_{i}x_{i}}}{\sum{w_{i}}}=\frac{\sum\limits_{i=1}^{n}{i(i^{2}+i)}}{\sum\limits_{i=1}^{n}{(i^{2}+i)}} $ $ =\frac{\sum\limits_{i=1}^{n}{i^{3}}+\sum\limits_{i=1}^{n}{i^{2}}}{\sum\limits_{i=1}^{n}{i^{2}}+\sum\limits_{i=1}^{n}{i}} $ $ =\frac{\frac{n^{2}{{(n+1)}^{2}}}{4}+\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}} $ $ =\frac{\frac{n(n+1)}{2}{ \frac{n(n+1)}{2}+\frac{2n+1}{3} }}{\frac{n(n+1)}{2}{ \frac{2n+1}{3}+1 }} $ $ =\frac{3n^{2}+7n+2}{2(2n+4)}=\frac{(3n+1)(n+2)}{4(n+2)}=\frac{3n+1}{4} $
BETA
